Arithemtic and Algebra problems from a handwritten notebook labeled *Gwynedd School 1807-1808 Scholars Names and Sundries,* used by the teacher, Benjamin Albertson.

1. Question

As I was beating on the forest grounds, up starts a hare before my two grey-hounds.

The dogs, being light of foot, did fairly run unto her fifteen rods just twenty-one.

The distance that she started up before was four-score sixteen rods just, and no more.

Now this I'd have you unto me declare: How far they ran before they caught the hare.

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1. Answer

Let x = the distance required

Thus as hound 21: hare 15 :

therefore hound x:(15x/21) = the distance the hare ran after starting, til caught

x = (15x/21) + 96 by the question

21x = 15x + 2016 when cleared of fractions

21x - 15x = 2016 by transposition

6x = 2016

x = 2016/6 = 336 rods, the distance as required

or as 6:21 therefore 96:336

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2. Question

A Lady was asked her age, who replied thus

My age if multiplied by three,

Two sevenths of that product trippled be,

The square root of two ninths of that is four,

Now tell my age or never see me more.

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2a. Answer solved algebraicly

Let thy age be denoted by x.

Then 3x - 2/7 = 6x/7

6x/7 times 3 = 18x/7

2/9 of 18x/7 = 36x/63

the square root of 36x/63 = 4 by the question

consequently 36x/63 =16

36x = 1008 by the clearing of the fractions

x = 1008/36 = 28 the age required

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2b. Answer solved arithmetically

Let us suppose she was 14.

2/7 of (14 times 3) is 12

2/9 of (12 times 3) is 8, but should have been 16 by the question (i.e. 4 squared)

So the first guess has an error of 8.

Let us suppose she is 21.

2/7 of (21 times 3) is 18.

2/9 of (18 times 3) is 12, but again should have been 16.

The error of the second guess is 4.

The difference in the error is 8-4 = 4.

14 times 4 = 56

21 times 8 = 168

The difference in the products is 168-56 = 112

Dividing 112 by 4 we get 28, the age required.

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3. Question

If "A" can do a piece of work alone in ten days and "B" in thirteen, set them both about it together in what time will it be finished?

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3a. Answer

Suppose the work to consist of 130 parts. Then A will perform 13 and B 10 parts, which added make 23 parts of the work performed by them both in one day.

As 23 parts is to 1 day, 130 is to 5 and 15/23 = 5 days, 7 hours, 49 13/23 minutes

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3b. Second Method to obtain this Answer

10 days:1 work :: 1 day:(1/10)work

13:1 :: 1 day:(1/13) work

thus (1/10)+(1/13)=(23/130) work in a day

as 23/130 of the work:1/1 days :: 1/1 work: 5 and 15/23 [answer]

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3c. Answer by Algebra

Let x be the time sought.

Thus as 10 days:1 work therefore x days . . x/10

And as 13 days: 1 work :: x days .. x/13

Hence x/10 = part done by A in x days

And x/13 = part done by B in x days

Consequently (x/10) + (x/13) must = 1 work

That is (13x/10)+x = 13 or 13x + 10x = 130 by clearing it of fractions

Therefore 23x = 130 or x = 130/23 = 5 and 15/23 as before

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4. Question

To wit - I have an oblong meadow, In which my horse I sometimes tether' Its side AB is fourscore feet, Its corners all right-angles meet, My horse I tie at corner B, With line as long as side BC. If you this line for radius take, And corner B the center make; You'll find the circle will divide, In equal parts the other side, AD -- My data now are shown, I pray you make the contents known. |
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4a. Answer by trigonometry

It is evident from the question that ABE is an angle of thirty degrees. Consequently BEA is 60 degrees.

Now say, as the sine of BEA 60 degrees is 9.93753

Is to AB ------------------------- 80 ft. = 1.90309

So is radius ---------------- 90 degrees = 10.0000

1.90309 + 10.0000 = 11.90309

So BE =92.376 feet = (11.09309-9.93753) = 1.96556

The area in square feet is 80 feet times 92.376 feet = 7390.08

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4b Answer by Algebra, Thus,

Let x = AE

Consequently BE must be equal to 2x

Then according to a rule in trigonometry, which says "The sum of the squares of the two legs of any plane triangle is equal to the square of the hypotenuse," we have the equation:

4x squared = x squared + 6400

3 x squared = 6400 by transposition

x squared = 2133 and one third

square root of 6400 + x squared = square root of 6400 + 2133 and 1/3 = square root of 8533 and 1/3

which = 92.376 feet which = BE.

When multiplied by 80 feet this gives the answer 7390.08 as before.

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5. Question

A gentleman did live, no matter where,

A wife and two small children, all his care.

Returning from his fields, in generous humour,

Of apples brought his hat-full, and a few more.

He gave his wife of his delicious store,

Half what he had, and half an apple more.

His eldest son, making a mighty grapple,

Seized half of what was left, and half an apple.

Halft what was left he gave his youngest son

And half an apple, leaving himself just one.

Though at his nice division, who may wonder,

Yet not one apple did he leave assunder

I here by Algebra do make known

The number, which this Gentleman brought in.

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5. Answer

Let x = the number he brought in.

Then x/2 - 1/2 = what was left after his wife

1/2 of (x/2 - 1/2) = (x/4 - 1/4) which after subtracting an additional 1/2 = x/4-3/4 = what was left after his eldest son.

1/2 of (x/4-3/4) = (x/8 - 3/8) which after subtracting an additonal 1/2 = (x/8 - 14/16) = what was left after the youngest son.

(x/8 - 14/16) = 1 by the question.

16x - 112 = 128 by clearing it of fractions

16 x =112 + 128 = 240

x = 240/16 = 15, the number required.

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